Given, $$f(x) = {{4{x^3} - 3{x^2}} \over 6} - 2\sin x + (2x - 1)\cos x$$

$$f'(x) = {{12{x^2} - 6x} \over 6} - 2\cos x + (2x - 1)( - \sin x) + \cos x(2)$$

$$ = (2{x^2} - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x$$

$$ = 2{x^2} - x - 2x\sin x + \sin x$$

$$ = 2x(x - \sin x) - 1(x - \sin x)$$

$$f'(x) = (2x - 1)(x - \sin x)$$

for $$x > 0,x - \sin x > 0$$

$$x < 0,x - \sin x < 0$$

for $$x \in ( - \infty ,0] \cup \left[ {{1 \over 2},\infty } \right),f'(x) \ge 0$$

for $$x \in \left[ {0,{1 \over 2}} \right],f'(x) \le 0$$

Hence, f(x) increases in $$\left[ {{1 \over 2},\infty } \right)$$.